Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=3-2 \cos t, y=-1+5 \sin t ; t \neq n \pi $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{5}{2}\cot t\); \(\frac{d^2 y}{dx^2} = \frac{-5}{4}\csc^3 t\).

Step 1: Differentiate x and y with respect to t

First, find the derivatives of the parametric equations with respect to the parameter \(t\). For \(x=3-2\cos t\), differentiate to get \(\frac{dx}{dt} = 2\sin t\). For \(y=-1+5\sin t\), differentiate to get \(\frac{dy}{dt} = 5\cos t\).

Step 2: Find dy/dx

Use the derivatives from Step 1 to find \(\frac{dy}{dx}\) by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\). This gives \(\frac{dy}{dx} = \frac{5\cos t}{2\sin t}\). Simplify this to \(\frac{dy}{dx} = \frac{5}{2}\cot t\).

Step 3: Differentiate dy/dx with respect to t

Find the derivative of \(\frac{dy}{dx} = \frac{5}{2}\cot t\) with respect to \(t\). This requires using the chain rule. The derivative of \(\cot t\) is \(-\csc^2 t\). Therefore, \(\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{5}{2}(-\csc^2 t)\).

Step 4: Find d²y/dx²

Use the derivative from Step 3 to find \(\frac{d^2 y}{dx^2}\). Divide \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\) by \(\frac{dx}{dt} = 2\sin t\). Thus, \(\frac{d^2 y}{dx^2} = \frac{\frac{5}{2}(-\csc^2 t)}{2\sin t}\). Simplify this to \(\frac{d^2 y}{dx^2} = \frac{-5}{4}\frac{1}{\sin^3 t}\). Thus, \(\frac{d^2 y}{dx^2} = \frac{-5}{4}\csc^3 t\).

Key Concepts

chain rule

The chain rule is a fundamental tool in calculus utilized for finding the derivative of composite functions. Imagine you have a function within another function. To differentiate the outer function with respect to an original variable, you use the chain rule. In our problem, we applied the chain rule when differentiating \( \frac{dy}{dx} \) with respect to \( t \). Here's why it's essential:

• In parametric equations, the variables x and y are often expressed in terms of a parameter, like t.
• To find derivatives with respect to x, we first differentiate with respect to t. This intermediary step often requires the chain rule.
• When differentiating expressions like \((\frac{dy}{dx})\), the chain rule helps manage the complexity.

The chain rule states that the derivative of two composed functions is the product of their derivatives. For instance, if you have functions f and g, such that y = f(g(t)), the chain rule is applied as \( \frac{dy}{dt} = f'(g(t)) \cdot g'(t) \). It offers a structured approach to tackle derivatives in complex, multi-layered functions. In our case, differentiating \((\frac{dy}{dx} = \frac{5}{2} \cot t)\) necessitates this rule, making the process of finding higher derivatives manageable.

differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate of change of the function with respect to a variable. It's a key concept in calculus and forms the basis for analyzing changes in mathematical models. In the given exercise:

• The primary goal is to find derivatives. Here, we start by finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), which are the rates of change of x and y with respect to t.
• These derivatives allow us to express \( \frac{dy}{dx} \) in terms of t, illustrating how y changes with x.
• Finally, we aim to find higher-order derivatives like \( \frac{d^2 y}{dx^2} \) which describe changes of changes, giving deeper insights into the curve's behavior.

Differentiation not only aids in understanding how variables relate, but also reveals critical points and concavities on a graph. In the case of parametric equations, differentiation helps traverse the step from parameter-based representations to concrete derivatives expressing the intrinsic relationships between x and y.

parametric equations

Parametric equations represent functions with a common variable, usually time or an angle, which is known as the parameter. Instead of expressing y as a direct function of x, both x and y are written as functions of another variable t. Here's what you need to know:

• Parametric equations like \( x=3-2\cos t \) and \( y=-1+5\sin t \) offer a powerful way to define curves, often simplifying complex shapes into more manageable forms.
• They allow for easier manipulation and view of the curve in two-dimensional space.
• The parametrization in terms of \( t \) particularly helps in understanding physical phenomena, where time is a fundamental aspect, like in motion paths.

Using parameter t, both x and y can be differentiated with respect to t to explore how the curve changes. By finding \( \frac{dy}{dx} \) or higher derivatives without eliminating t, we keep the detailed interplay between x and y intact, revealing the dynamics embedded in the parameters. This approach efficiently manages the description of shapes and motions that can't easily be captured using just y in terms of x.